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Question
Solve the previous problem if the friction coefficient between the 2⋅0 kg block and the plane below it is 0⋅5 and the plane below the 4⋅0 kg block is frictionless.
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Solution
From the figure, we have
\[m_2 gsin\theta - T_2 = m_1 a..........(1)\]
\[ T_1 - \left( m_1 g\sin\theta + \mu m_1 g\cos \theta \right) = m_1 a ............(2)\]
\[ T_2 - T_1 = \frac{Ia}{r^2} .............(3)\]
Adding equations (1) and (2), we get
\[m_2 g\sin\theta - \left( m_1 g\sin\theta - \mu m_1 g\cos \theta \right) + \left( T_1 - T_2 \right) = m_1 a + m_2 a\]
\[m_2 g\sin\theta - \left( m_1 g\sin\theta + \mu m_1 g\cos \theta \right) - \frac{Ia}{r^2} = m_1 a + m_2 a\]
\[ m_2 g\sin\theta - \left( m_1 g\sin\theta + \mu m_1 g\cos \theta \right) = m_1 a + m_2 a + \frac{Ia}{r^2}\]
\[4 \times 9 . 8 \times \frac{1}{\sqrt{2}} - \left\{ 2 \times 9 . 8 \times \frac{1}{\sqrt{2}} + 0 . 5 \times 2 \times 9 . 8 \times \frac{1}{\sqrt{2}} \right\} = \left( 4 + 2 + \frac{0 . 5}{0 . 01} \right)a\]
\[ \Rightarrow 27 . 80 - 13 . 90 + 6 . 95 = 56a\]
\[ \Rightarrow 27 . 8 - 20 . 8 = 56a\]
\[ \Rightarrow a = \frac{7}{56} = 0 . 125 m/s^2\]
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